Bearing Capacity (Daya Dukung Tanah)


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1 Bearing Capacity (Daya Dukung Tanah) Dr. Ir.H. Erizal, MAgr
2 Definisi Daya dukung yang diizinkan (allowable bearing cap.) tekanan maksimum yang dapat diaplikasikan ke tanah dimana 2 kondisi diatas dipenuhi. Daya dukung batas (ultimate bearing cap.) tekanan minimum yang menyebabkan keruntuhan geser (shear failure) pada tanah pendukung secara cepat ke bawah.
3 Footing Performance Vertical Load Vertical movement Elastic maximum tolerable settlement safe load Plastic Serviceability Ultimate Limit State Maximum allowable load = min [safe load, max service load ] maximum service load Plunging Failure ultimate capacity
4 Limit States Serviceability Ultimate
5 Serviceability Limit State Maximum load at which structure still performs satisfactorily : Settlement Horizontal movement Rotation Sliding Applied Load Force (kn)
6 Bearing Pressure Definitions Allowable Bearing Pressure q a = < qf s (settlement) / A F Plan Area, A Ultimate Bearing Capacity q ult = F fail / A
7 Foundation Failure Rotational Failure Force Soil Heave Resistance
8 Generalized Shear Failure q Soil Failure Lines Settlement rigid radial shear passive log spiral
9 Local Shear Failure q minor surface heave only Settlement Medium dense or firm soils
10 Punching Shear Failure q No surface heave Settlement Loose or Soft Soils
11 Methods for calculating bearing capacity Full scale load tests Load tests on model footings Limit equilibrium analysis Detailed stress analysis such as the FEM method
12 Limit equilibrium analysis solutions for weightless soils: Solutions with φ = 0 : Prandtl smooth punch : q ult = 5.14c Prandtl rough punch : q ult = 5.7c Solutions with φ 0 : Rough punch passive active log spiral
13 Bearing Capacity for real soils Exact, theoretical analytical solutions have only been computed for special cases  e.g. soils with no weight, no frictional strength, φ or no cohesion, c. Approximate solutions have been derived by combining solutions for these special cases. The first solution was by Terzaghi (1943)  father of soil mechanics. Others later modified this solution. The failure mechanism corresponds to general failure. Corrections are applied to check for the possibility of local or punching shear failure.
14 Jenis pondasi berdasarkan kedalamannya 1. Pondasi dangkal (shallow foundation) bila kedalaman pondasi, Df, lebih kecil dibanding lebar pondasi, B 2. Pondasi dalam (deep foundation) bila kedalaman pondasi, Df, lebih besar/dalam dibanding lebar pondasi, B
15 Terzaghi s Bearing Capacity Eqn. For strip footings: q ult = c.n c + σ ZD.N q + 0.5γ BN γ φ c
16 Terzaghi s Bearing Capacity Eqn. For strip footings: q ult = c.n c + q.n q + 0.5γ BN γ φ c D f q = γ.d γ.d f B soil density, γ (kn/m 3 )
17 Terzaghi s Bearing Capacity Eqn. For strip footings: q ult = c.n c + q.n q + 0.5γ BN γ Bearing Capacity Factors for soil cohesion, surcharge and weight functions of friction angle, φ determine by equation or from graph
18 40 N c N q N γ Ø in Degrees N c and N q N γ
19 General Bearing Capacity Eqn. (1973, 1975) Based on theoretical and experimental work: q ult = c.n c F cs F cd F ci + q.n q F qs F qd F qi + 0.5γBN γ F γs F γd F γi φ c
20 General Bearing Capacity Eqn. q ult = c.n c F cs F cd F ci + q.n q F qs F qd F qi + 0.5γBN γ F γs F γd F γi φ c D f q = γ.d σ ZD = f γ.d B soil density, γ (kn/m 3 )
21 General Bearing Capacity Eqn. q ult = c N c F cs F cd F ci + qn q F qs F qd F qi + 0.5γBN γ F γs F γd F γi Bearing Capacity Factors for soil cohesion, surcharge and weight functions of friction angle, φ determine by equation or from graph or Table 3.3
22 General Bearing Capacity Eqn. q ult = c N c F cs F cd F ci + qn q F qs F qd F qi + 0.5γBN γ F γs F γd F γi Correction factors for footing shape (s), footing depth (d) load inclination (i ); could have additional base inclination (b), and ground inclination (g) determine from appropriate equations
23 General Bearing Capacity Factors (Table 3.3) Nγ Hansen 40 Friction angle (degree) Nγ Meyerhof Nc 10 Nq Nc, Nq and Nγ
24 Wall on Strip Footing Shape Factors Bird s s Eye View Column on Square Footing For nonstrip footings : F cs, F cq, F γs 1 Failure lines Failure lines
25 Wall on Strip Footing Depth Factors For buried footings : F cd, F qd, F γd 1 q = γ.d f increased strength failure generally line length increases with depth
26 V = kn H = 423 kn Inclination Factors For inclined loads : F ci, F qi, F γi 1 Inclined load = 1000 kn Load inclination, θ = 25 o Failure surface shallower and shorter
27 Terzaghi or General General is more accurate Applies to a broader range of loading and geometry conditions General is more complicated
28 Contoh 1 Sebuah pondasi bujur sangkar dengan sisi 2.25 m diletakkan pada kedalaman 1.5 m pada pasir< di mana parameter kuat gesernya c =0 dan ø= 38 o. Tentukan daya dukung ultimit (a) bila muka air tanah berada di bawah elevasi pondasi, (b) jika muka air tanah berada pada permukaan tanah. Berat isi pasir di atas muka air tanah adalah 18 kn/m 3, berat isi jenuhnya 20 kn/m 3. Pondasi bujur sangkar q f = 0.4γBN γ + γdn q ø= 38 o N γ = 67, N q = 49 q f = (0.4 x 18 x 2.25 x 67) + (18 x 1.5 x 49) = = 2408 kn/m 2 Daya dukung di bawah muka air: q f = 0.4γ BN γ + γ DN q γ = γ sat γ w = = 10.2 kn/m 3 q f = (0.4 x 10.2 x 2.25 x 67) + (10.2 x 1.5 x 49) = = 1365 kn/m 2
29 Contoh 2 Sebuah pondasi jalur didesain memikul beban 800 kn/m pada kedalaman 0.7 m pada pasir berkerikil. Parameter kekuatan geser yang tersedia adalah c =0 danø =40 o. Tentukan lebar pondasi bila faktor keamanan = 3 dan diasumsikan mungkin muka air tanah mencapai pondasi. Berat isi pasir adalah 17 kn/m 3, berat isi jenuhnya 20 kn/m 3. ø =40 o N γ =95 dan N q =64 q f = ½γ BN γ + γbn q = (½ x 10.2 x B x 95) + (17 x 0.7 x 64) = 485B q nf =q f γd ; q n = q  γd ; F = q nf / q n = 485B (17 x 0.7) = (800/B) (17 x 0.7) = 485B = (800/B) 12 1 (485B + 750) = B B = 1.55 m
30 Ultimate Bearing Capacity of Shallow Footings with Concentric Loads
31 Ultimate Bearing Capacity with Ground Water Effect
32 Example: Determine the Allowable Bearing Capacity for A Rough Base Square Footing Using A Safety Factor Of 3. d = D = 5 γ T = 125 pcf B = 6 γ sub = 63 pcf φ = 20 c = 500 psf
33 Solution: Assuming A General Shear Condition, Enter the Bearing Capacity Chart for φ= 20 and Read N c = 14, N q = 6, Nγ = 3. Also note that formula for bearing capacity must account for the square footing and the water table within the failure zone. q ult B = ( )CNc + [ γ subd + ( γt γsub )d]nq + 0.4γ subbnγ L = ( 1.3)(500)14 + [63(5) + (125 63)5] (63)(6)(3) = q ult = 13,300psf q all = q ult 3 = 13, , 430 psf
34 What is the Effect on Bearing Capacity of Excavation of Soil Cover Over a Spread Footing?
35 Student MiniExercise on Bearing Capacity q ult = cn c + P o N q + 1/2 γ BN γ Properties and Dimensions (Assume Continuous Rough Footing) γ = Unit Weight D = Footing Embedment B = Footing Width Cohesive Soil φ = 0 c = 1000psf q ult (psf) Cohesionless Soil φ = 30 c = 0 q ult (psf) A. Initial Situation γ T = 120 pcf, D = 0, B = 5, deep water table B. Effect of embedment D = 5, γ T = 120 pcf, B = 5, deep water table C. Effect of width, B = 10, γ T = 120 pcf, D = 0, deep water table D. Effect of water table at surface, γ sub = 57.6 pcf, D = 0, B = 5
36 Student MiniExercise on Bearing Capacity q ult = cn c + P o N q + 1/2 γ BN γ Properties and Dimensions (Assume Continuous Rough Footing) γ = Unit Weight D = Footing Embedment B = Footing Width Cohesive Soil φ = 0 c = 1000psf q ult (psf) Cohesionless Soil φ = 30 c = 0 q ult (psf) A. Initial Situation γ T = 120 pcf, D = 0, B = 5, deep water table B. Effect of embedment D = 5, γ T = 120 pcf, B = 5, deep water table C. Effect of width, B = 10, γ T = 120 pcf, D = 0, deep water table D. Effect of water table at surface, γ sub = 57.6 pcf, D = 0, B =
37 Footing Bearing Capacity Objective: STUDENT EXERCISE NO.5 Find the Allowable Bearing Capacity Using a Safety Factor = 3, for the Condition Shown Below. Rough Base Footing Final Grade Sand γ = 115 pcf φ = 35 C = 0
38 Length Width 50 Footing = = 5 > 9 Water Level Width 30 4 = = SOLUTION TO EXERCISE No. 5 Use Rectangular Formula No Water Effect = 2.6 > 1.5 Footing Widths below Footing Base = (115)(4)(37) + (0.4)(115)(10)(42) = 17, ,320 = 36,340 PSF 10 qult = γ DNq + 0.4γ BN γ Qall 36,340 = = 12,113 psf 3
39 How is bearing capacity theory related to the rule of thumb equation for stability; SAFETY FACTOR = 6 C γ H Soft clay layer Compact Sand H γ = Unit Weight cohesion = C
40 Spread Footing Design Bearing Capacity Explain how footing embedment, width, and water table affect footing bearing capacity Activities: Bearing capacity analysis
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